Answer:
Microphones are a type of transducer - a device which converts energy from one form to another. Microphones convert acoustical energy (sound waves) into electrical energy (the audio signal). Different types of microphone have different ways of converting energy but they all share one thing in common: The diaphragm.
Explanation:
14. Which of the following is an example of changing acceleration?
a. A car stopped at a red light.
b. A boat sailing west at 5 knots.
c. A train traveling 65 miles per hour south.
d. A man jogging 3 meters/second along a winding path.
Fiona claims that the diagram below shows simple machines, but Chad claims that it shows a compound machine.
Answer:
Simple machine
Explanation:
It is a simple machine because the person wants to raise the load by an inclined plane. Simple machines perform work with the mechanical advantage offered by the machine itself, such as using a bar as Lever, lifting a load by means of a pulley. simple machines multiply or change direction a force. While Composite machines are the Union of several simple machines that perform a given job, examples of combined machines are found on bicycles, a washing machine, a car, and others.
Answer:
b
Explanation:
took test, other answer said b and was verified
A 12.0 V battery is attached to a circuit with a resistor of 5.5 Ohms. The internal resistance of the circuit is 1.75 Ohms. What is the terminal voltage of the circuit?
The end voltage is 9.01 volts.
Explanation:The emf of the battery is 12 volts.
The resistance of the circuit = 1.75 ohms.
The resistance of the applicant attached = 5.5 ohms.
So, total resistance of the circuit = [tex]1.75 + 5.5[/tex] ohms = 7.25 ohms.
According to the equation, the voltage applied is equal to the product of the current and the resistance.
V=iR.
So, 12 = [tex]i \times7.25[/tex].
So, i = 1.66.
Therefore, the voltage at the end of the circuit = [tex]12 - 1.66\times1.75[/tex]volts
= 9.01volts.
So the end voltage is 9.01 volts.
How much pressure is applied to the ground
by a 93 kg man who is standing on square
stilts that measure 0.04 m on each edge?
Answer in units of Pa.
003 (part 2 of 2) 10.0 points
What is this pressure in pounds per square
inch?
Answer in units of lb/in.
Answer:
Pressure applied by the man= 285103.125 [tex]Pa[/tex] or 41.35 [tex]lb/in^{2}[/tex]
Explanation:
Pressure is defined as the perpendicular force applied per unit area.
i.e. [tex]Pressure=\frac{Force}{Area}[/tex]
Now, [tex]Force= mg[/tex]
where, [tex]m[/tex] = mass of the body(man) = 93 kg
[tex]g[/tex] = acceleration due to gravity of Earth = 9.81 [tex]m/{s^{2}}[/tex]
[tex]Area[/tex] covered is equal to the area of both stilts(a man generally stands on two feet)
therefore [tex]Area=2(0.04)^{2}[/tex] [tex]m^{2}[/tex]
and putting in the values, we get,
[tex]Pressure=\frac{93\times9.81}{2\times(0.04)^{2}}Nm^{-2}=285103.125Nm^{-2}[/tex]
Now we need to convert to our required units:
[tex]1Nm^{-2}=1Pa\\1Pa=0.000145038lb/in^{2}[/tex]
(We can get the above result by individually converting kg to lb and meters to inches respectively)
Using the above relations we get,
[tex]Pressure=285103.125Pa=0.000145038\times285103.125lb/in^{2}=41.35lb/in^{2}[/tex]
Based on the formula for pressure, the pressure applied on the ground is 570206.25 N/m² or 82.7 pound per square inch
What is pressure?Pressure is defined as the perpendicular force applied per unit area.
Pressure = Force/AreaAlso,
Force = mgwhere,
mass of the body = 93 kg
g = 9.81 m/s²
Force = 93 * 9,81
Force = 912.33 N
Area of square stilt = (0.04m)²
Area = 1.6 * 10⁻³ m²
Then;
Pressure = 912.33 N/1.6 * 10⁻³ m²
Pressure = 570206.25 N/m²
Converting to pounds per square inch:
1 pound per square inch = 6894.76 N/m²
570206.25 N/m² = 570206.25 N/m² * 1 pound per square inch/6894.76 N/m²
Pressure = 82.7 pound per square inch
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A fully loaded Saturn V rocket has a mass of 2.92 x 106 kg. Its engines have a thrust of 3.34 x 107 N.
Complete Question:
A fully loaded Saturn V rocket has a mass of 2.92 x 106 kg. Its engines have a thrust of 3.34 x 107 N. (8 marks)
a) What is the downward force of gravity on the rocket at blast-off?
b) What is the unbalanced force on the rocket at blast-off?
c) What is the acceleration of the rocket as it leaves the launching pad?
d) As the rocket travels upwards, the engine thrust remains constant, but the mass of the rocket decreases. Why?
e) Does the acceleration of the rocket increase, decrease, or remain the same as the engines continue to fire?
Answer:
a) [tex]-2.8616 \times 10^{7} N[/tex] is the downward force of gravity on the rocket at blast-off.
b) [tex]4.784 \times 10^{6} N[/tex] is the unbalanced force on the rocket at blast-off
c) [tex]1.638 \mathrm{m} / \mathrm{s}^{2}[/tex] is the acceleration of the rocket as it leaves the launching pad
d) Because the propellant here is burned up, hence the mass of the rocket seems to be varied (total mass of all its parts). Thereby, the mass decreases when the rocket moves upward.
e) The acceleration of the rocket increases when engines continue to fire
Explanation:
Given:
Mass (m) = [tex]2.92 \times 10^{6} \mathrm{kg}[/tex]
a) In physics, weight can be defined as the applied force on a body by gravity. It is the product of mass (m) and gravity [tex]\left(g=9.8 \mathrm{m} / \mathrm{s}^{2}\right)[/tex]
[tex]\text { weight }(W)=m \times g=2.92 \times 10^{6} \times(-9.8)=28.616 \times 10^{6}=-2.8616 \times 10^{7} N[/tex]
The negative sign indicates the downward force of gravity.
b) To find the unbalanced force on the rocket at blast-off,
Accelerating force,
[tex]F_{a}=F+W=3.34 \times 10^{7}+\left(-2.8616 \times 10^{7}\right)=(3.34-2.8616) \times 10^{7}[/tex]
[tex]F_{a}=0.4784 \times 10^{7}=4.784 \times 10^{6} N[/tex]
c) Newton’s second law of motion states that the object’s acceleration depends on two variable:
Directly proportionate to the object’s force existed Inversely proportionate to the mass of the objectsThe equation can be given as below,
[tex]Force =m \times acceleration[/tex]
[tex]\text { Acceleration }=\frac{F_{a}}{m}=\frac{4.784 \times 10^{6}}{2.92 \times 10^{6}}=1.638 \mathrm{m} / \mathrm{s}^{2}[/tex]
d) The pushing of rocket upward will happen as long as the engine gets fired. The propellant here is burned up, hence the mass of the rocket seems to be varied (total mass of all its parts). Thereby, the mass decreases (taotal mass) when the rocket moves upward.
e) The acceleration of the rocket increases when engines continue to fire
Let consider [tex]F_{a}[/tex] is constant, mass gets decreasing, then the acceleration would be increasing (as mass and acceleration are inversely proportionate to each other) .
3. What is the net force acting on the box in the diagram below? (1 mark)
Answer: Fnet = 40 N on the right
Explanation: The net force is the summation of forces acting on an object. Add the forces in the same direction and subtract to the opposing force moving in the opposite side.
Fnet= F1 + F2 - F3
= 120 N + 60 N - 140 N
= 180 N - 140 N
= 40 N on the right
A battery is marked. As having a voltage of 12V and is connected to a circuit that has a resistance of 5 ohms.if the terminal voltage is 8.75V. What is the internal resistance
Answer:
1.857 Ohms
Explanation:
The marked voltage of the battery is it's E.M.F. (Electromotive force). It is the maximum voltage the battery can supply because, ideally internal resistance of a cell or battery should be zero. Let us denote the E.M.F. as E.
In reality we get a lower voltage called the Terminal Voltage (Voltage across the terminals of the battery). Let us call that V. This is due to the voltage drop caused by the internal resistance (say 'r').
So when the circuit is connected a current I flows through a resistance
R(= 5 ohms, given).
Now, potential difference = Terminal Voltage = V = 8.75 Volts
Hence, [tex]I=\frac{V}{R} =\frac{8.75}{5}=1.75 Amperes[/tex]
Now, by conservation of energy we can say that,
E.M.F. = Terminal Voltage + Voltage drop due to the battery's internal resistance
or, [tex]E=V+Ir[/tex]
or, [tex]r=\frac{E-V}{I}=\frac{12-8.75}{1.75}=1.857[/tex] Ohms
a 3kg has 9 joules of kinetic energy how fast is it flying
Answer: 2.4 m/s
Explanation:use KE= 1/2 mv²
Derive to find v:
v = √ 2KE / m
= √ 2 ( 9 J ) / 3 kg
= 2.4 m/s
5) Consider pushing a 50.0 kg box through a 5.00m displacement on both a flat surface and up a
ramp inclined to the horizontal by 15.0°. In both cases, you apply a force of 100. N parallel to the
surface (parallel to the floor or parallel to the ramp). Calculate the work done by:
a) the gravitational force as the box is pushed across the flat ground
b) the gravitational force as the box is pushed up the ramp
c) the force you apply as the box is pushed across the flat ground
d) the force you apply as the box is pushed up the ramp
Explanation:
Work equals the force times the parallel distance.
a) Force gravity is in the downward direction. The box is moving on flat ground, so there's no displacement in that direction.
W = Fd
W = (50 kg) (-9.8 m/s²) (0 m)
W = 0 J
b) This time, there is a displacement in the vertical direction.
W = Fd
W = (50 kg) (-9.8 m/s²) (5.00 m sin 15.0°)
W = -634 J
c) The force is parallel to the displacement.
W = Fd
W = (100. N) (5.00 m)
W = 500. J
d) Again, the force is parallel to the displacement.
W = Fd
W = (100. N) (5.00 m)
W = 500. J
1. Suppose you give a shopping cart a push and it begins t
give a shopping cart a push and it begins to roll across the parking
lot. Once you have stopped pushing it, it slows down and comes to a sto
observation consistent with Newton's First Law? Why or why not?
Newton's first law says the cart should keep going if there's no force acting on it to slow it down. But it slows down and stops. Is there any force acting on it ? You bet ! There's friction in the wheels, and a little bit of air resistance too.
A short circuit is _____.
Answer:
in a device, an electrical circuit of lower resistance than that of a normal circuit, typically resulting from the unintended contact of components and consequent accidental diversion of the current.
Explanation:
Answer: A short circuit is simply a low resistance connection between the two conductors supplying electrical power to any circuit. This results in excessive current flow in the power source through the 'short,' and may even cause the power source to be destroyed.
Explanation:
Which statement describes Newton’s conception of the Solar System?
1.) the Sun at the center, with the Earth and other planets orbiting the Sun
2.) the Earth orbiting the Sun, with other planets orbiting the Earth
3.) the Earth at the center, with the Sun and other planets orbiting Earth
4.) the Sun orbiting the Earth, with other planets orbiting the Sun
PLEASE HELP I NEED THIS ANSWER QUICK!
Answer:
1). the sun at the center, with the earth and other planets orbiting the sun.
Explanation:
Newton believes that it's gravity between the sun and the other planets even at the large distances that cause them to orbit around the sun. This force build up by the masses of the sun and the planets which keeps the planets in their orbit.
Answer:the sun at the center, with the earth and other planets orbiting the sun.
Explanation:
What effect does Earth’s spherical shape have on the amount of sunlight that reaches the equator?
A.
The latitudes near the equator receive direct sunlight, which causes a high-pressure system.
B.
The latitudes near the equator receive indirect sunlight, which causes a low-pressure system.
C.
The latitudes near the equator receive direct sunlight, which causes a low-pressure system.
D.
The latitudes near the equator receive indirect sunlight, which causes a high-pressure system.
A
Explanation:
The earth is spherical. So it's middle part is bulgjng outside. So more sunlight will be incident on the latitude near the equator. This will heat the air and it will rise up. This will cause high pressure difference and polar disturbances.
I hope this satisfies you. I hope u will follow me and make this the brainliest answer.
The Earth's spherical shape ensures that the equator receives the most direct sunlight, resulting in higher temperatures and a low-pressure system due to the rising warm air. This leads to the high precipitation in the equatorial region and plays a significant role in global atmospheric and oceanic circulation patterns.
The spherical shape of the Earth influences the way sunlight is distributed across different latitudes. At the equator, the sun's rays are most direct, resulting in a higher concentration of solar energy and higher temperatures. This intense sunlight is focused on a smaller area, leading to the warm conditions that are characteristic of equatorial regions. Due to the heat at the equator, the air warms up, becomes less dense and creates a low-pressure system.
As a consequence of the heat and low-pressure conditions, there is significant convection, which causes air to rise. This is part of the reason why the equatorial region, also known as the Intertropical Convergence Zone, is often characterized by high precipitation levels. The rising warm air can lead to cloud formation and frequent rainfall. In contrast, at higher latitudes, the sun's rays strike the surface at a more oblique angle, spreading the energy over a wider area and resulting in cooler temperatures and different atmospheric conditions.
Therefore, the correct answer to the question is: C. The latitudes near the equator receive direct sunlight, which causes a low-pressure system.
what is 1/12+7/9 hj
Explanation:
[tex] \frac{1}{12} + \frac{7}{9} \\ \\ = \frac{1 \times 3}{12 \times 3} + \frac{7 \times 4}{9 \times 4} \\ \\ = \frac{3}{36} + \frac{28}{36} \\ \\ = \frac{3 + 28}{36} \\ \\ = \frac{31}{36} \\ \\ \huge \purple{ \boxed{\therefore \: \frac{1}{12} + \frac{7}{9} = \frac{31}{36} }} \\ [/tex]
Answer:
31/36
Explanation:
do the math you will see
4) Consider a separate rocket, also in deep space with a mass of 30.0 kg. If the rocket is
observed to be travelling at v= 5.00 m/s at t= 3.00 s and then travelling at û = -2.00 m/s at t= 14.0 s
with constant acceleration, calculate:
a) the acceleration, a.
b) the force F, acting on the rocket from the thrusters,
Answer:
(a) -0.636 m/s²
(b) -19.08 N
Explanation:
Given:
Mass of the rocket (m) = 30.0 kg
Initial velocity of rocket (v) = 5.00 m/s
Initial time of rocket (t₁) = 3.00 s
Final velocity of the rocket (u) = -2.00 m/s
Final time of rocket (t₂) = 14.0 s
(a)
Acceleration is given as the rate of change of velocity. Therefore,
[tex]a=\frac{u-v}{t_2-t_1}\\\\a=\frac{-2.00-5.00}{14.00-3.00}\ m/s^2\\\\a=\frac{-7.00}{11.00}\ m/s^2\\\\a=-0.636\ m/s^2[/tex]
Therefore, the acceleration of the rocket is 0.636 m/s².
(b)
From Newton's second law, we know that, force acting on a body is equal to the product of its mass and acceleration.
So, the force acting on the rocket is given as:
[tex]F=ma\\\\F=(30.0\ kg)(-0.636\ m/s^2)\\\\F=-19.08\ N[/tex]
Therefore, the force acting on the rocket is -19.08 N.
The negative sign implies the force acts in the direction opposite to motion.
A 10-newton box rests on a table. What is the net force on the box? A. 0 N B. 10 N C. more than 10 N D. between 0 N and 10 N
A. 0 N is the net force acting on the box
Explanation:
Newton's first law of motion gives an insight to this problem, if an object stays in its position of either motion or rest, it can only change its current of either motion or rest is by applying an external force on it. If an object is resting on a table, there is a normal force acting perpendicular to the object and at the same, there is the gravitational force acting on the object in the perpendicular direction opposite to the direction of the normal forceHence, these two forces cancel and the resultant force will be zero. so it is in rest and it moves only when the resultant force is non zero.Generators may be set up to produce either direct or _____ current
Answer:
Generators may be set up to produce either direct or alternating current
Explanation:
What are five of the most important things you learned in physics? ASAP
Answer:
The definitions of mass, energy, time and space as used in Physics are circular
The Conservation of Mass and Energy
The definition of zero is not sufficiently defined in Physics
Newton’s Laws of Motion are related by Calculus
Knowing the terms of Physics is most important.
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two different examples of forces that could cause a change in motion
Answer:
a)Kicking a ball; b) A car that gets hit
Explanation:
a)
A classic example is when a soccer player kicks a moving ball in another direction by modifying the original direction and speed that the ball brought.
b)
Another example is when a car goes on a highway and suddenly is hit by another car on its side modifying the original direction and speed.
She sights two sailboats going due east from the tower. The angles of depression to the two boats are 42o and 29o. If the observation deck is 1,353 feet high, how far apart are the boats?
Answer:
The boats are 934.65 feet apart
Explanation:
Given:
The angles of depression to the two boats are 42 degrees and 29 degrees
Height of the observation deck i = 1,353 feet
To Find:
How far apart are the boats (y )= ?
Solution:
Step 1 : Finding the value of x(Refer the figure attached)
We can use the tangent ratio to find the x value
[tex]tan(42^{\circ}) = \frac{1353}{x}[/tex]
[tex]x = \frac{1353}{tan(42^{\circ}) }[/tex]
x = 590.47 feet
Step 2 : Finding the value of z (Refer the figure attached)
[tex]tan(29^{\circ}) = \frac{1353}{z }[/tex]
[tex]z = \frac{1353}{tan(29^{\circ})}[/tex]
z = 1525.12 feet
Step 3 : Finding the value of y (Refer the figure attached)
y = z -x
y = 1525.12 - 590.47
y = 934.65 feet
Thus the two boats are 934.65 feet apart
An object is placed 12cm from a converging lens of focal length 18cm. Find the position of the image.
Answer:
The position of the image = 7.2 cm.
Explanation:
Given:
A Converging lens:
Object distance [tex](d_o)[/tex] = 12 cm = -12 cm (sign convention)
Focal length [tex](f)[/tex] = 18 cm
We have to find the position of the image.
Let the position of the image be [tex](d_i)[/tex] .
Sign convention:
The focal length of converging (convex) lens is always positive,while object distance is negative.
Using lens formula:
⇒ [tex]\frac{1}{object\ distance} + \frac{1}{image\ distance} =\frac{1}{focal\ length}[/tex]
⇒ [tex]\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}[/tex]
⇒ [tex]\frac{1}{d_i} =\frac{1}{f} - \frac{1}{d_o}[/tex]
⇒ [tex]\frac{1}{d_i} =\frac{1}{f} - \frac{1}{-d_o}[/tex]
⇒ [tex]\frac{1}{d_i} =\frac{1}{f} + \frac{1}{d_o}[/tex]
⇒ [tex]d_i=\frac{d_o\times f}{d_o+f}[/tex]
⇒ [tex]d_i=\frac{12\times 18}{12+18}[/tex]
⇒ [tex]d_i=\frac{216}{30}[/tex]
⇒ [tex]d_i=7.2[/tex] cm
So the position of the image = 7.2 cm.
A lump of steel of mass 10kg at 627 degree Celsius is dropped in 100kg oil at 30 degree Celsius . the specific heat of steel And oil are 0.5kj/kg.k and 3.5kj/kg.k calculate the entropy change in steel,oil and in the universe.
Answer:
700J
Explanation:
What are the results of inserting a crimp connector into the crimp tool facing the wrong way ?
If you insert a crimp pin incorrectly, the ratcheted crimp tool will not sufficiently crimp the tabs. As a result, the wire may not fully conduct with the pin and the pin will be damaged.
Explanation:
The general theory for crimping all types of connectors is to strip a little bit of insulation off the wire. Then, put the connector into a suitably sized space in the jaws, insert the wire, and crimp it down. For non-ratcheting pliers, it's suggested the connector be re-crimped with the next smallest hole in the jaws.
A good crimp connection is gas tight and won't wick: it is sometimes referred to as a “cold weld”. Like the solder method, it can be used on solid or stranded conductors, and provides a good mechanical and electrical connection.
Inserting a crimp connector into a crimp tool incorrectly can lead to poorly fastened connections, resulting in signal loss, intermittent connectivity, or failure of the electrical connection and may create a safety hazard.
Inserting a crimp connector into a crimp tool facing the wrong way can result in a poor connection as the connector will not be properly fastened to the wire. The specific results depend on the type of crimp connector and tool, but generally, you may end up with a connector that is not adequately compressed, leading to signal loss, intermittent connectivity, or complete failure of the electrical connection. Additionally, it increases the risk of short-circuiting or creating a potential safety hazard since the contact between the conductor and the terminal may not be secure.
Light in a vacuum travels at a constant speed of 3x10^8 m/s. If the moons average distance from the earth is 38776106 km how long would it take for a beam of light to travel from earth to the moon and back to earth again?
Answer: 258.3 s
Explanation:
The speed [tex]s[/tex] is given by the following equation:
[tex]s=\frac{D}{t}[/tex]
Where:
[tex]s=3(10)^{8} m/s[/tex] is the speed of light in vacuum
[tex]D=2(38776106 km \frac{1000 m}{1 km})=7.75(10)^{10} m[/tex] is the double of the distance between Earth and Moon, since the beam of light travels from Earth to the Moon and back to Earth again.
[tex]t[/tex] is the time it takes to the beam of light to travel the mentioned distance
Isolating [tex]t[/tex] and solving with the given information:
[tex]t=\frac{D}{s}[/tex]
[tex]t=\frac{7.75(10)^{10} m}{3(10)^{8} m/s}[/tex]
Finally:
[tex]t=258.3 s \approx 258 s[/tex]
If you were able to jog one mile in 10 minutes, how would you progressively increase your
performance by using each of the following? Be specific by using numbers in your answers.
1. (F) frequency variable?
2. (1) intensity variable?
3. (T) time variable?
Answer:
1. (F) Increasing the frequency variable reduces the time required.
2. (I) Intensity variable can be related the the energy expended.
3.(T) Time variable would affect the speed.
Explanation:
1. (F) increasing the frequency variable implies that the distance would be covered in lesser time. As a wave of 250Hz cover more distance than that of 200Hz.
2. (I) Intensity variable. An increase in this variable implies that more energy is expended.
3.(T) Time variable can either give the required out come or not. As increasing the time would mean that the speed has reduced, while decreasing the time means the distance would be covered quickly. If the required speed is 20m/s, then increasing it to 30m/s imples that lesser time would be recorded. If the reverse is the case, then more time would be recorded. time = distance/speed
1. reduces the time required.
2.energy expended.
3.affect the speed.
What is Frequency, time period and intensity ?Frequency, time period and intensity are quantitative dimensions which describe the amount of physical activity taking place during a given time.
1. (F) frequency variable - Frequency refers to the number of waves that pass a fixed point in unit time.
i.e. [tex]F =\frac{1}{T}[/tex]
When frequency is increased then distance would be covered in lesser time.
For example, a wave of 550Hz cover more distance than that of 500Hz.
2. (I) Intensity variable- The quantity of energy the wave conveys per unit time across a surface of unit area.
i.e. [tex]I = \frac{P}{4\pi r^{2} }[/tex]
When Intensity increases it implies that more energy is expended.
3.(T) Time variable -It is define as the ratio of distance and time.
i.e. [tex]time = \frac{distance}{speed}[/tex]
When time variable is increased than the quantity of energy the wave conveys per unit time across a surface of unit area, speed is reduced.
When time variable is decreased the distance would be covered quickly.
For example : If the speed of car is 80m/s, then by increasing the speed to 100m/s shows that lesser time would be required.
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A force of 30N acts through a distance of 4m in the direction of the force, what is the work done
Answer:
120J
Explanation:
Work done = Force (N) X Distance (m)
Answer:
Work = 120Nm or 120J
Explanation:
Work is said to be done when a force moves a body over a distance, in the direction of force applied.
Work done = force x distance (in the direction of the force)
Therefore work = force x displacement
As displacement is distance in a specific or specified direction.
From the parameters given
Force = 30N
Displacement = 4m.
Therefore work = force x displacement
= 30N x 4m
Work = 120Nm or 120J
Note :J is joules which is the S. I unit of work
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10,500 J of GPE, weight 539 N, how tall is the hill you are sitting on?
To calculate the height of the hill, use the formula h = GPE/Weight with the provided gravitational potential energy (10,500 J) and weight (539 N) of the object. The height equals the GPE divided by the weight, which, upon calculation, gives the height of the hill in meters.
The student is asking how to calculate the height of the hill, given the gravitational potential energy (GPE) and the weight of an object. In physics, the formula to calculate GPE is GPE = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s2 on Earth or 10 m/s2 for simplified calculations), and h is the height.
To find the height, one can rearrange the GPE formula to solve for h, which gives us h = GPE/(mg). The weight of an object is the force due to gravity acting on it, which can be calculated by multiplying the mass by the acceleration due to gravity (Weight = mg). Given that the weight (Weight) of the object is known, we can simplify the formula for height to h = GPE/Weight.
Using the information provided by the student, the GPE is 10,500 J, and the weight is 539 N. Substituting these values into our height formula gives us h = 10,500 J / 539 N, which results in an answer after calculation.
Let's calculate the height:
GPE = 10,500 J
Weight = 539 N
Height h = GPE / Weight = 10,500 J / 539 N
=19.48
A 0.250 kg mass is attached to a spring with k=18.9 N/m. At the equilibrium position, it moves 2.89 m/s. What is the amplitude of the oscillation? (Unit=m)
Answer: y = R*sin(2.75*t - δ)
Here δ is just the time offset and for our purposes is pretty irrelevant. You can in fact set it to zero since we can say we begin timing when the mass crosses equilibrium. So
y = R*sin(2.75*t)
We want to find a way to use the information "At the equilibrium position, it moves 2.89 m/s." I am going to use some calculus here since it makes things so much easier. If you haven't taken calculus yet, most likely your course has given you a formula to use instead.
We know y=0 when t=0, so y is at equilibrium when t=0. To say it moves 2.89 m/s is then to say that
y'(0) = 2.89.
From here we can differentiate the displacement function, set t=0 and solve for R. Using the chain rule:
y'(t) = 2.75*R*cos(2.75*t)
y'(0) = 2.75*R
2.75*R = 2.89
R = 1.051
Explanation: Since this is harmonic motion we can assume there is no damping force. The frequency of the oscillation is given by ω=√(k/m)=√(18.9/2.5)=2.75. Keep in mind this is angular frequency, i.e. radians per second, not wavelengths per second.
How does slope on this graph indicate the amount of density?
Density comparison
Question 3 options:
A higher density is indicated by a slope that is not steep (yellow line)
The line in the middle (red line) has the highest density
The steeper slope (green line) indicates a higher density
Final answer:
The slope of a line on the graph represents the amount of density. A steeper slope indicates a higher density, while a flatter slope indicates a lower density.
Explanation:
The slope of a line on the graph represents the amount of density in this scenario. A steeper slope indicates a higher density, while a flatter slope indicates a lower density.
For example, in the given graph, the steeper slope (green line) indicates a higher density, while the slope of the yellow line is not steep, indicating a higher density as well.
Therefore, the statement 'The steeper slope (green line) indicates a higher density' is correct.
67. Race Car A race car is slowed with a constant accel-
eration of 11 m/s2 opposite the direction of motion.
a. If the car is going 55 m/s, how many meters will it
travel before it stops?
b. How many meters will it take to stop a car going
twice as fast?
Answer:
a. 137.5 m
b. 550 m
Explanation:
Accelerated Motion
If an object is changing its velocity at a constant rate, it has a uniformly accelerated motion. When the object is moving in one fixed axis, then the sign of the acceleration is negative if the object is braking, and positive if the object is increasing its speed.
The initial speed vo, final speed vt, acceleration a, and distance traveled x are related by the formula
[tex]v_f^2=v_o^2+2ax[/tex]
a. The Race Car A has an initial speed of 55 m/s and it's said to stop. We must find at what distance it goes to vf=0. This means that the above formula becomes
[tex]0=v_o^2+2ax[/tex]
Solving for x
[tex]\displaystyle x=-\frac{v_o^2}{2a}[/tex]
The acceleration is [tex]-11\ m/s^2[/tex], negative because it's against the movement. Thus
[tex]\displaystyle x=-\frac{55^2}{2\times (-11)}[/tex]
[tex]x=137.5\ m[/tex]
b. If the car is going twice as fast (v0=110 m/s), then
[tex]\displaystyle x=-\frac{110^2}{2\times (-11)}[/tex]
[tex]x=550\ m[/tex]